Question

The vertex of the graph of a quadratic functon is (3,1) and the y intercept is (0,10). What is the equations of the function?

Answer 1

\[\text{Let } f(x) = k(x-a)^2 + c\] (general quadratic with square completed)

(3,1) is a minimum => a = .... and c = ....
When x = 0, f(x) = 10, so k = ...

We can now expand and find the solution in the form

f(x) = px^2 + qx + r

Answer 2

No idea what to do after that. But thank you!

Answer 3

Well, if a quadratic is at a minimum, x = a in the bracket (as the bracket is >= 0 for all x)

So the only value is the + c

This occurs at the point x = 3, y = 1, so we can deduce that the "a" = 3, and the c = 1

=> f(x) = k(x-3)^2 + 1

When x = 0, f(x) = 10. So k* (-3)^2 + 1 = 10 -> 9k + 1 = 10 -> k = 1

So f(x) = (x-3)^2 + 1, which we could expand (if we want) to f(x) = ....

Answer 4

¬_¬

Answer 5

I didn't get it correct :/

Answer 6

What answer did you give?

Answer 7

(x-3)^2 + 1 = x^2 - 6x + 10, which is the answer

Answer 8

I don't remember! :/ it's an onlne class so I can't go back and change my anwers.